How do you find the antiderivative of #tan^2(x) dx#

Answer 1

Evelyn Agard

The answer is: #tanx-x+c#.

Remembering that the derivative of #y=tanx# is #y'=1+tan^2x#,

Than:

#inttan^2xdx=int(tan^2x+1-1)dx=#

#=int(tan^2x+1)dx-intdx=tanx-x+c#.

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Answer 2

Scarlett Allison

You can start by writing #tan^2(x)=sin^2(x)/cos^2(x)# giving: #inttan^2(x)dx=intsin^2(x)/cos^2(x)dx=#

using: #sin^2(x)=1-cos^2(x)# you get:

#=int(1-cos^2(x))/(cos^2(x))dx=int[1/cos^2(x)-1]dx=# #=int1/cos^2(x)dx-int1dx=#

#=tan(x)-x+c#

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Answer 3

Charles Arocha

To find the antiderivative of ( \tan^2(x) , dx ), you can use integration by parts or trigonometric identities. One common approach is to use the identity ( \tan^2(x) = \sec^2(x) - 1 ) and rewrite the integral in terms of secant:

[ \int \tan^2(x) , dx = \int (\sec^2(x) - 1) , dx ]

[ = \int \sec^2(x) , dx - \int dx ]

[ = \tan(x) - x + C ]

Where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.